Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $r = \dfrac{z^2 + 11z + 30}{9z^2 + 27z} \times \dfrac{z + 3}{-8z - 48} $
First factor the quadratic. $r = \dfrac{(z + 6)(z + 5)}{9z^2 + 27z} \times \dfrac{z + 3}{-8z - 48} $ Then factor out any other terms. $r = \dfrac{(z + 6)(z + 5)}{9z(z + 3)} \times \dfrac{z + 3}{-8(z + 6)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (z + 6)(z + 5) \times (z + 3) } { 9z(z + 3) \times -8(z + 6) } $ $r = \dfrac{ (z + 6)(z + 5)(z + 3)}{ -72z(z + 3)(z + 6)} $ Notice that $(z + 3)$ and $(z + 6)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ \cancel{(z + 6)}(z + 5)(z + 3)}{ -72z(z + 3)\cancel{(z + 6)}} $ We are dividing by $z + 6$ , so $z + 6 \neq 0$ Therefore, $z \neq -6$ $r = \dfrac{ \cancel{(z + 6)}(z + 5)\cancel{(z + 3)}}{ -72z\cancel{(z + 3)}\cancel{(z + 6)}} $ We are dividing by $z + 3$ , so $z + 3 \neq 0$ Therefore, $z \neq -3$ $r = \dfrac{z + 5}{-72z} $ $r = \dfrac{-(z + 5)}{72z} ; \space z \neq -6 ; \space z \neq -3 $